A Game on Noetherian Rings

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A friend suggested the following combinatorial game. At any time, the state of the game is a (commutative) Noetherian ring $\neq 0$. On a player's turn, that player chooses a nonzero non-unit element of the ring, and replaces the ring with its quotient by the ideal generated by that element. The player to make the last legal move wins, by passing the opponent a field.

So if the ring was $\mathbb C[x,y]/(x^2+y^2-1)$, and a player chooses $x$, the ring becomes $\mathbb C[y]/(y^2-1)$. This is a poor move, as his opponent can turn the ring into a field by choosing either $y+1$ or $y-1$, and win. A winning move would have been $x+iy+1$, which turns the ring into a field immediately and wins the game.

Problem: Given a ring, how can we tell if it is a winning position for the first player or for the second player?

(The game terminates since the original ring is Noetherian, and an unending game would be an infinite ascending chain in the original ring.)

Will Sawin

Posted 2012-04-06T03:24:54.820

Reputation: 66 112

63Does it tell too much about me that I think this game would be kind of fun? – Olivier – 2012-04-06T08:06:44.923

3Uhm..nice! I hoped to reduce to regular rings, so the winning position would coincide with the value $\pmod{2}$ of the Krull dimension. But since quotients of regular rings may fail to be regular again (see $\mathbb{Z}/p^2$), it leads nowhere... – Filippo Alberto Edoardo – 2012-04-06T09:12:34.833

1@FAE: I don't believe that Krull dimension is the right thing to consider. In a geometric situation like that of the OP, the minimal dimension of an irreducible component should play a role (you can win in one move if you have the union of a line and a plane for instance). And there are reduction issues too (start with a double line). – Jérôme Poineau – 2012-04-06T10:37:32.973

3I am absolutely shocked no one has pointed out the similarities to Choquet's game. Baire's theorem gives a characterization of the existence of a winning strategy for one of the players: www.math.auckland.ac.nz/~moors/game.pdf

Question: Why not use limites to extend the game to any type of ring? – Malte – 2012-04-06T11:40:16.073

2@Jerôme: I absolutely agree with you, I was reporting on a bad idea... – Filippo Alberto Edoardo – 2012-04-06T12:00:21.680

5Question: are the algebra softwares strong enough so that it would possible to actually implement this game ? If so, we could organized a tournament. That would be really fun. – Joël – 2012-04-06T18:56:19.623

1Does anyone have a idea if $k[x,y]$ wins or loses? – Martin Brandenburg – 2012-05-04T14:25:02.023

If $k$ is algebraically closed, the player who plays on it wins. Any Weierstrauss equation will do, since the opponent must pass you a ring of finite $\geq 2$ dimension over $k$, with which you can always pass them a field. If it's not algebraically closed, this might not work, and I'm not sure what to do. My guess is that if $k$ is a number field then the first player to play loses. – Will Sawin – 2012-05-04T19:56:16.353

In fact, Hilbert irreducibility implies that this is true. – Will Sawin – 2012-05-04T19:58:03.017

@Will: Thanks; but I don't understand the argument. I don't see why every ring of finite vector space dimension $\geq 2$ over $k$ can be moved to some field immediately, i.e. that it has some principal maximal ideal. – Martin Brandenburg – 2012-05-09T19:36:04.123

The ring is a quotient of a Dedekind domain, thus, a product of quotients of DVRs. Choose the maximal ideal from one DVR and the unit ideal from the others.

Also: In my Hilbert irreducibility argument, I neglected to mention the obviously critical assumption that the original polynomial is irreducible. Playing something reducible, like x(x-1), forces them to pass you a finite-dimension non-field (with a principal maximal ideal.) – Will Sawin – 2012-05-09T19:47:36.683

1This is also similar to Sylver Coinage. – Gabriel C. Drummond-Cole – 2014-08-20T16:39:49.490

1@Joël: I have written a GAP-program which helps to analyse the game for finite rings. One has to input the initial ring using the method RingByStructureConstants. – Martin Brandenburg – 2017-05-05T08:15:15.367

Answers

41

I computed the nimbers of a few rings, for what it's worth. I don't see any sensible pattern so perhaps the general answer is hopelessly hard. This wouldn't be surprising, because even for very simple games like sprouts starting with $n$ dots no general pattern is known for the corresponding nimbers.

OK so the way it works is that the nimber of a ring $A$ is the smallest ordinal which is not in the set of nimbers of $A/(x)$ for $x$ non-zero and not a unit. The nimber of a ring is zero iff the corresponding game is a second player win -- this is a standard and easy result in combinatorial game theory. If the nimber is non-zero then the position is a first player win and his winning move is to reduce the ring to a ring with nimber zero.

Fields all have nimber zero, because zero is the smallest ordinal not in the empty set. An easy induction on $n$ shows that for $k$ a field and $n\geq1$, the nimber of $k[x]/(x^n)$ is $n-1$; the point is that the ideals of $k[x]/(x^n)$ are precisely the $(x^i)$. In general an Artin local ring of length $n$ will have nimber at most $n-1$ (again trivial induction), but strict inequality may hold. For example if $V$ is a finite-dimensional vector space over $k$ and we construct a ring $k\oplus \epsilon V$ with $\epsilon^2=0$, this has nimber zero if $V$ is even-dimensional and one if $V$ is odd-dimensional; again the proof is a simple induction on the dimension of $V$, using the fact that a non-zero non-unit element of $k\oplus\epsilon V$ is just a non-zero element of $V$, and quotienting out by this brings the dimension down by 1. In particular the ring $k[x,y]/(x^2,xy,y^2)$ has nimber zero, which means that the moment you start dealing with 2-dimensional varieties things are going to get messy. But perhaps this is not surprising -- an Artin local ring is much more complicated than a game of sprouts and even sprouts is a mystery.

Rings like $k[[x]]$ and $k[x]$ have nimber $\omega$, the first infinite ordinal, as they have quotients of nimber $n$ for all finite $n$. As has been implicitly noted in the comments, the answer for a general smooth connected affine curve (over the complexes, say) is slightly delicate. If there is a principal prime divisor then the nimber is non-zero and probably $\omega$ again; it's non-zero because P1 can just reduce to a field. But if the genus is high then there may not be a principal prime divisor, by Riemann-Roch, and now the nimber will be zero because any move will reduce the situation to a direct sum of rings of the form $k[x]/(x^n)$ and such a direct sum has positive nimber as it can be reduced to zero in one move. So there's something for curves. For surfaces I'm scared though because the Artin local rings that will arise when the situation becomes 0-dimensional can be much more complicated.

I don't see any discernible pattern really, but then again the moment you leave really trivial games, nimbers often follow no discernible pattern, so it might be hard to say anything interesting about what's going on.

Kevin Buzzard

Posted 2012-04-06T03:24:54.820

Reputation: 27 668

Nice answer. The OP seems to be asking a simpler question, though: Is there an algorithm for computing the nimber of a given Noetherian ring? Obviously this will depend on how the ring is specified, but I assume that you managed to figure out an algorithm for certain kinds of specifications? – Timothy Chow – 2012-04-08T03:14:00.543

7The OP is asking "is the nimber zero or not" -- but the only algorithm I know for figuring out whether the nimber is zero or not is easily modified to one which computes the nimbers anyway, and is of the form "compute what's going on with all the quotients". The algorithm for computing a nimber is recursive -- you compute nimbers of all quotients by principal ideals and then take the mex. I just ran this algorithm for some simple Artin rings. Even for Artin rings the answer will be very complicated I should think. – Kevin Buzzard – 2012-04-08T07:57:57.203

@Kevin: By your remarks, the problem is also interesting for one-dimensional domains, for which you can sometimes give an answer (with examples with nimber zero or not) without fully solving the Artinian case. – YCor – 2012-04-08T18:28:23.003

1

@Kevin: Thanks for the additional comment. By the way, it's true in surprisingly great generality that determining whether or not the nimber is zero is no easier than computing the nimber itself. See "Nimbers are inevitable," http://arxiv.org/abs/1011.5841

– Timothy Chow – 2012-04-09T03:13:07.323

@Kevin: Your reasoning for $k[x]$ only shows that the nimber is $\geq \omega$, right? For equality, we have to show that the proper quotients $k[x]/(f)$ have nimber $<\omega$. For this, one shows by induction on $\mathrm{deg}(f)$ that the nimber is $\leq \mathrm{deg}(f)-1$. – Martin Brandenburg – 2012-04-19T16:18:18.097

Right -- I was implicitly invoking the fact that a ring for which the game only has a finite number of plays, has finite nimber. – Kevin Buzzard – 2012-04-19T18:51:25.087

Oh, I understand: If $R$ is a finite-dim vector space over $k$, then its nimber is $<\omega$. – Martin Brandenburg – 2012-04-20T06:16:15.693

Am I right in thinking that by taking path algebras with relations that say two paths are equal that you can realise any impartial Hackenbush game as the game of a Noetherian ring (which is Artinian iff the Hackenbush graph is finite)? – Bruce Westbury – 2012-04-21T17:18:50.433

It sounds to me like a hard problem to produce a Noetherian ring corresponding to an arbitrary game tree such that all paths through it are finite. Do you want to be more explicit about the "by taking path algebras with relations" bit?? – Kevin Buzzard – 2012-04-21T18:15:21.657

I am not starting from an arbitrary game nor am I looking at a game tree. I am starting from an impartial Hackenbush position. – Bruce Westbury – 2012-04-24T16:04:53.327

OK. Can you be more explicit about the "by taking path algebras with relations" bit? – Kevin Buzzard – 2012-04-25T06:44:51.523

I can try. Let me formulate it differently. There is a variation of this question with Noetherian categories. A move consists of nominating a morphism and taking the quotient by the ideal it generates. This seems like playing Hackenbush with directed graphs. You could now allow a move to specify an arbitrary ideal and get a different game. You could then take the algebra of the category and play the game in the question. A fourth variation is to start with a Noetherian ring and a move consists in passing to the quotient by a principal ideal. – Bruce Westbury – 2012-04-25T19:33:03.583

@Bruce: So you want to construct a ring on a rooted graph where there is a bijection between ideals and sets of edges closed upwards, with principal ideals being the ones generated by a particular edge? – Will Sawin – 2012-04-27T02:58:22.647

1This seems problematic because one should then be able to divide a higher edge by a lower edge, and get a non-invertible number that does not clearly correspond to any edge. – Will Sawin – 2012-04-27T02:59:59.273

@Kevin: Could you explain your statement "because any move will reduce the situation to a direct sum of rings of the form $k[x]/(x^n)$"? – Martin Brandenburg – 2012-05-03T08:07:13.693

The $k$ in my assertion is not the ground field $k$, it can be bigger. Sorry for the confusion. I'm just using that any ideal is the product of prime ideals and the chinese remainder theorem. It doesn't matter for the purposes of this question anyway -- it's a far stronger assertion than what is necessary. – Kevin Buzzard – 2012-05-05T08:25:01.367

One gets better formulas if one includes the zero ring; for example then $k[x]/(x^n)$ has nimber $n$, and this is even better if we have products of such rings. See also Section 5.4 of my revised paper https://arxiv.org/abs/1205.2884

– Martin Brandenburg – 2017-05-05T07:37:07.077

29

Let us call a Noetherian commutative ring $R$ an $\mathcal{N}$-position if the next player has a winning strategy; otherwise the previous player has a winning strategy and we will call it a $\mathcal{P}$-position. The zero ring is allowed and hence we will choose the misère play rule, i.e. the player with the last move loses (see Tom Goodwillie's comment). For example, it is clear that the zero ring is $\mathcal{N}$, that fields are $\mathcal{P}$, and that PIDs which are no fields are $\mathcal{N}$.

In general, $R$ is $\mathcal{P}$ iff $R/\langle x \rangle$ is $\mathcal{N}$ for all $0 \neq x \in R$, and $ R$ is $\mathcal{N}$ iff either $R=0$ or $R/\langle x \rangle$ is $\mathcal{P}$ for some $0 \neq x \in R$. As a general theme, $\mathcal{P}$-positions are quite rare and hard to find, and every $\mathcal{P}$-position is responsible for many $\mathcal{N}$-positions which are then more easy to find.

The following results are proven here:

  • If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Rem. 5.1).
  • Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$; otherwise, $R$ is $\mathcal{P}$ (Prop. 5.6).
  • It follows that $K[X,Y]/\langle f \rangle $ is $\mathcal{P}$ when $f$ is a Weierstrass equation and $K$ is an algebraically closed field (Prop. 5.7).
  • One can also show that the coordinate ring of the cusp $K[X,Y]/\langle Y^2-X^3 \rangle$ is $\mathcal{P}$ (Prop. 5.13).
  • Hence, $K[X,Y]$ is $\mathcal{N}$ if $K$ is algebraically closed.
  • This actually holds for every field $K$, because $K[X,Y]/\langle X^2 \rangle$ is $\mathcal{P}$ (Prop. 5.16).
  • For this one needs that $K[X,Y]/\langle X^2,f^{n+1},f^n X \rangle$ is $\mathcal{P}$ for every irreducible $f \in K[Y]$ (special case of Prop. 5.15).

I conjecture that $K[X_1,\dotsc,X_n]$ is $\mathcal{N}$ for all $n \geq 1$.


We can also play such a game with groups. One starts with a group $G$. A move consists in replacing $G$ by $G/\langle\langle a \rangle\rangle$, i.e. we quotient out the smallest normal subgroup containing some element $a \neq 1$. The ending condition holds iff the ascending chain condition on normal subgroups holds (do these groups have a name?). When $G$ is abelian, this means that $G$ is finitely generated.

Actually we can play this game for every algebraic structure: We start with an algebra $A$ of a given signature. A move consists in replacing $A$ by $A/(a \sim b)$, where $a,b \in A$ with $a \neq b$.

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the mentioned article. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ under both play rules:

If $A$ is a finitely generated abelian group, then

  • $A$ is a normal $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some abelian group $B$.
  • $A$ is a misère $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.

Martin Brandenburg

Posted 2012-04-06T03:24:54.820

Reputation: 32 755

Interesting. If we restrict to abelian $p$-groups, then the game with $\mathbf{Z}/p^{a_1} \mathbf{Z} \oplus \ldots \oplus \mathbf{Z}/p^{a_n} \mathbf{Z}$ seems to be equivalent to Nim with $(a_1,\ldots,a_n)$. – François Brunault – 2012-04-22T09:24:44.203

1@Francois: No since we can also mod out, say, $(1,1,\dotsc)$. – Martin Brandenburg – 2012-04-22T09:38:30.137

10

The idea of a hierarchy on Noetherian rings as suggested earlier is good, but it doesn't reflect the structure of the problem. For example, a move need not reduce the "type" of a ring (it is clear from definition however, that a move cannot reduce the type by more than 1). For instance, the ring $\mathbb{C}[x,y] / (x^2 + y^2-1)$ is of "type 1" (as noted in the original post). If we mod out by $x$, we obtain the ring $\mathbb{C}[y]/(y^2 - 1)$, which is also of "type 1".

It is also not clear (at all!) that an optimal move "ought" to decrease the "type" by 1 if possible.


Here is a different idea that resolves this issue and creates a hierarchy of Noetherian rings preserving a "win/lose" structure:

First, slightly change the original game so that you are allowed to mod out by a unit element and the win conditions become "the first player to pass his opponent the zero ring loses" [this game is the same as the original one].

Then we construct the following rooted directed acyclic graph, $G$:

  • The vertices of $G$ are the "set" of Noetherian rings, where two rings are the same vertex iff they are isomorphic [to even discuss this collection of isomorphism classes requires the axiom of choice, and it's probably too big to be called a set, but I don't think that's too important].

  • The directed edges of $G$ are you draw an edge from $R$ to $S$ iff there is some $0 \neq r \in R$ such that $S \cong R/(r)$ [i.e., iff a player can get to $S$ from $R$ in one move].

Then the graph $G$ is acyclic and rooted (with root $0$) since any path in this graph must have finite length terminating in 0 (since all the rings are Noetherian).

Finally, to analyze the game, we just need to mark each vertex as "win for player 1" or "win for player 2" in the usual way.


Things to consider

  • I believe the "type" of a ring, $R$, as defined earlier is just the length of the shortest path from $R$ to a field.

  • To what extent can we algorithmically determine (pieces of) the graph $G$?

  • If we were magically given the graph $G$, to what extent can we use it to algorithmically determine who wins in each case?

-Pat Devlin

Pat Devlin

Posted 2012-04-06T03:24:54.820

Reputation: 1 622

If $R$ is a PID then the problem has a complete solution.

Case I: If $R$ is a field, player 1 loses.

Case II: If $R$ is not a field, let $I \subseteq R$ be a maximal proper ideal. Then since $R$ is a PID, $I=(r)$ for some $0 \neq r \in R$. But then $R/I = R/(r)$ is a field [because $I$ was a maximal ideal]. Therefore, with this move for player 1, he can make it so that player 2 loses.

I do not know if this can be extended to $R$ a UFD or not. – Pat Devlin – 2012-04-06T16:50:57.780

1I doubt it can be extended to a UFD, since UFDs can have quotients that are not UFDs. – Will Sawin – 2012-04-06T18:08:20.290

1It might be nice to consider a generalization of the ideal class group that measures how far a ring is from being a PID. Then you could hope this group is finite, and that the parity of its cardinality solves the problem... just a guess, though. – Philip van Reeuwijk – 2012-04-07T10:52:33.313

3If we are interested in the game starting from a particular ring $R$, the relevant part of $G$ can be represented as follows: vertices are the ideals of $R$, and there is an edge from $I$ to $J$ if $J=I+aR$ for some $a\notin I$. In any case, it seems to me that this is just a trivial restatement of the problem (any game can be represented as a graph). Is there anything the upvoters see and I am missing? – Emil Jeřábek – 2012-04-07T13:33:32.973